# Wednesday January 8

Reference: [@silverman_2009]

## Summary

1. Mordell-Weil theorem

  - For elliptic curves over global fields 
    - Number fields, function fields, finite fields, etc.
  
  - Proof uses Galois cohomology and height functions.
    Essentially one proof!
  
  - Holds for abelian varieties, but more difficult.
  - Need an analog of height functions, i.e. an $x\dash$coordinate).

2. Height functions.

3. Elliptic curves over $\QQ_p$ or complete discrete valuation fields[^silv_ref_one], particularly Tate curves.

4. Weil-Chatelet groups $E/k$ related to $H^1(k; E)$ with coefficients in the elliptic curve

5. Galois representation of $E/k$ for $\ch k = 0$, for 
  \[
  \rho_n: g_k \to \GL(2, \ZZ/n\ZZ)
  \]
  which leads to 
  \[
  \hat \rho: g_k \to \GL(\hat \ZZ)
  \]

[^silv_ref_one]:
See Silverman for basics, possibly Chapter 5

## Mordell-Weil Groups

Let $E/k$ be an elliptic curve over a field $k$[^note_on_fields], i.e. a smooth, projective, geometrically integral curve of genus 1 with a $k\dash$rational point $O$.

[^note_on_fields]:
Silverman is good for foundations, but assumes $k$ is a perfect field.
Here we'll let $k$ be arbitrary.

:::{.remark}
If $k$ is not algebraically closed, such a point $O$ may not exist.

By Riemann-Roch (easy computation) $E$ embeds (non-canonically) into $\Bbb{P}^2/k$ as a Weierstrass cubic 
\begin{align*}
y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6 \quad \Delta \neq 0
.\end{align*}

This is a smoothness condition, and this equation has a $k\dash$rational point at infinity $[0: 1: 0]$.
The line at infinity is a flex line (?), and so only intersects this curve at one point.
If $\ch k \neq 2,3$ then $y^2 = x^3 + Ax + B$.
Every elliptic curve is given by a Weierstrass equation, although not in a unique way.
:::


:::{.fact title="An amazing one!"}
The set of $k\dash$rational points $E(k)$ form an abelian group with zero as the identity.
:::

:::{.proof title="?"}
\envlist

1. Given any plane cubic $C/k$ and an origin $O \in C(k)$, the chord and tangent process yields a group structure.
  Note that there is a symmetry in connecting rational points $a, b$ with a line an intersecting at another rational point $c$ which is not present in most groups, so an additional inversion about $O$ is needed to actually make this into a group. 
  Proving associativity: difficult!

2. Look at $\mathrm{Pic}^0 E$, the degree 0 divisors on $E$ mod birational equivalence (?), which is equal to the degree 0 line bundles on $E$ mod bundle isomorphism.


:::{.exercise title="?"}
Show there is a map $C(k) \to \mathrm{Pic}^1 C$ given by sending $p$ is its equivalence class; this is a bijection by Riemann-Roch (straightforward exercise).
:::

We can then compose this with a map $\mathrm{Pic}^1 \to \mathrm{Pic}^0 C$ given by $D \mapsto D - [O]$, which decreases the degree by 1.
This gives a map $\Phi: C(k) \to \mathrm{Pic}^0 C$, just need to check that $\Phi(P \oplus Q) = \Phi(P) + \Phi(Q)$.
Check that the groups are independent of the $k\dash$rational point chosen, i.e. changing rational points yields isomorphic groups.
So the group law itself **does** actually depend on the rational point, although the structure doesn't.
:::

:::{.exercise title="?"}
Let $(E, O)/k$ be an elliptic curve and define $E^0 = E\setminus \theset{0}$ the (nonsingular, integral) affine curve given by removing the point at infinity.
Then the affine coordinate ring $k[E^0]$ is defined as $k[x, y]/(y^2 -x^3 - Ax - B)$, which is a Dedekind ring.

> The interesting thing about Dedekind domains: the ideal class group! (i.e. the Picard group)

This has ideal class group $\mathrm{Pic} k[E^0]$, and one can show that

\begin{align*}
\pic^0 E &\to \pic k[E^0] \\
\sum_p n_p \deg(p) [p] &\mapsto \sum_{p\neq 0} n_p [p] = \prod_p p^{n_p}
\end{align*}

with the sum ranging over all closed points is an isomorphism.

> Just note that the RHS can't have a point at infinity, so we just forget it.
> The isomorphism follows from some exact sequence with correction terms that vanish.

So the Mordell-Weil group of $E(k)$ is isomorphic to $\pic k[E^0]$, the class group of a dedekind domain (?).
:::

:::{.definition title="Class Group and the Mordell-Weil Group"}
Let $G$ be a commutative group.

- $G$ is a **class group** iff there exists a dedekind domain $R$ such that $G \cong \pic R$.
- $G$ is an **(elliptic) Mordell-Weil group** iff there exists a field $k$ and an elliptic curve $E/k$ such that $G \cong E(k)$.
:::

**Questions**:

1. Which $G$ are class groups?
2. Which $G$ are Mordell-Weil groups?

**Answer 1:**

:::{.theorem title="Clayborn, 1966"}
Every commutative $G$ is a class group.[^comm_alg_ref_1]
:::


[^comm_alg_ref_1]: 
Subsequent proofs: Leetham-Green (1972) and Clark (2008) following Rosen, and uses elliptic curves. See the end of Pete's Commutative Algebra notes!

**Answer 2**:
Consider $E/\CC$, then $E(\CC) \cong S^1 \cross S^1$, so the torsion subgroup is 
$$
T(1) \definedas (\QQ/\ZZ)^2 = \bigoplus_{\ell} (\QQ_\ell/\ZZ_\ell)^2
.$$
This in fact holds for any algebraically closed field of characteristic zero.

:::{.fact}
For any $E/k$, the Mordell-Weil group $E(k)$ is "$T(1)$-constrained", i.e. $E(k)[\mathrm{tors}] \injects T(1)$.
:::

:::{.theorem title="Clark, 2012"}
$G$ is a Mordell-Weil group $\iff G$ is $T(1)\dash$constrained.
:::

:::{.remark title="Some open problems."}
The analogous statement for abelian varieties, i.e being $T(g)$ constrained for some other genus $g\neq 1$, is open.
Fixing $k= \QQ$ still yields very interesting problems. 
Computing the rank and torsion subgroups is currently open, and the subject of modern research.
:::